91Ó°ÊÓ

Let \(y = \sin^{-1}x\). Then, \(x = \sin{y}\). To find \(\frac{dy}{dx}\), we will differentiate \(x = \sin{y}\) with respect to \(x\) using implicit differentiation. Differentiating both sides with respect to \(x\) gives: \(\frac{d}{dx}(x) = \frac{d}{dx}(\sin{y})\) \(1 = (\cos{y})\frac{dy}{dx}\) To solve for \(\frac{dy}{dx}\), we need to express \(\cos{y}\) in terms of \(x\). Using the Pythagorean identity \(\sin^2{y} + \cos^2{y} = 1\), we have: \(\cos^2{y} = 1 - \sin^2{y}\) Substitute \(sin^2y\) using \(x = \sin y\): \(\cos^2{y} = 1 - x^2\) Take the square root of both sides: \(\cos{y} = \sqrt{1 - x^2}\) Now substitute this expression for \(\cos{y}\) in our equation for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{1}{\cos{y}} = \frac{1}{\sqrt{1 - x^2}}\) So, the derivative of \(\sin^{-1}x\) is \(\frac{1}{\sqrt{1 - x^2}}\).

Let \(y = \cos^{-1}x\). Then, \(x = \cos{y}\). To find \(\frac{dy}{dx}\), we will differentiate \(x = \cos{y}\) with respect to \(x\) using implicit differentiation. Differentiating both sides with respect to \(x\) gives: \(\frac{d}{dx}(x) = \frac{d}{dx}(\cos{y})\) \(1 = (-\sin{y})\frac{dy}{dx}\) To solve for \(\frac{dy}{dx}\), we need to express \(\sin{y}\) in terms of \(x\). Using the Pythagorean identity \(\sin^2{y} + \cos^2{y} = 1\), we have: \(\sin^2{y} = 1 - \cos^2{y}\) Substitute \(\cos^2{y}\) using \(x = \cos y\): \(\sin^2{y} = 1 - x^2\) Take the square root of both sides: \(\sin{y} = \sqrt{1 - x^2}\) Now substitute this expression for \(\sin{y}\) in our equation for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{-1}{\sin{y}} = \frac{-1}{\sqrt{1 - x^2}}\) So, the derivative of \(\cos^{-1}x\) is \(\frac{-1}{\sqrt{1 - x^2}}\).

Now that we have the derivatives of both \(\sin^{-1}x\) and \(\cos^{-1}x\), let's compare them: Derivative of \(\sin^{-1}x: \frac{1}{\sqrt{1 - x^2}}\) Derivative of \(\cos^{-1}x: \frac{-1}{\sqrt{1 - x^2}}\) Looking at these expressions, we can see that the derivatives are related in the following way: \(\frac{d}{dx}(\cos^{-1}x) = -\frac{d}{dx}(\sin^{-1}x)\) In other words, the derivative of \(\cos^{-1}x\) is the negative of the derivative of \(\sin^{-1}x\).